Find $\lim_{x\to 4}\dfrac{x^3-4x^2}{x^2-16}$. Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $1$ (Choice C) C $-4$ (Choice D) D The limit doesn't exist
Substituting $x=4$ into $\dfrac{x^3-4x^2}{x^2-16}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression on our hands, let's try to simplify it. $\dfrac{x^3-4x^2}{x^2-16}$ can be simplified as $\dfrac{x^2}{x+4}$, for $x\neq 4$. This means that the two expressions have the same value for all $x$ -values (in their domains) except for $4$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x^3-4x^2}{x^2-16}=\dfrac{x^2}{x+4}$ for all $x$ -values in the interval $(3,5)$ except for $x=4$. Therefore, $\lim_{x\to 4}\dfrac{x^3-4x^2}{x^2-16}=\lim_{x\to 4}\dfrac{x^2}{x+4}=2$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 4}\dfrac{x^3-4x^2}{x^2-16}=2$.